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I Physical Meaning of complex wavenumber?

  1. May 17, 2016 #1
    Hey,
    This is my first post so I am hoping to do everything right :-)

    I do not understand the physical meaning of a complex wavenumber. I understand that, with a general approach u(x,t) = Re(A*[e][(i(kx-omega*t)]) and a complex wavenuber that the wave is decaying exponentially with x. What happens to the energy in that case? In the literature I can only find that the rod (wave propagation in solids) is absorbing all the energy. What kind of energy is it? Thermal energy due to friction (but that is not modeled) or destrucitve interferences (But where is it reflected?)

    Thanks for your response,
    Jens
     
  2. jcsd
  3. May 17, 2016 #2

    blue_leaf77

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    Photon's energy, it gets absorbed by the medium as the light passes through.
    The medium doesn't have to have the form of a rod. Any medium generally absorbs light to some extent.
    Interference may happen during light propagation but it doesn't affect the light's total energy.
     
  4. May 17, 2016 #3
    The OP did not say what kind of wave is he discussing. Maybe a sound wave or EM wave.
    The mechanism of absorption depends on the type of wave and the medium. Even for sound waves alone there are several mechanisms for energy dissipation.
    In the end it is all heat, but the path to get there may be different.
     
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